Question: $\begin{cases}b(1)=-7\\\\ b(n)=b(n-1)+12 \end{cases}$ Find the $4^{\text{th}}$ term in the sequence.
Solution: This is a recursive formula. It tells us that the first term is $-7$ and that the common difference is $12$. $\begin{aligned} {b(1)}&=-7 \\\\ {b(2)}&={b(1)}+12=5 \\\\ {b(3)}&={b(2)}+12=17 \\\\ {b(4)}&={b(3)}+12=29\end{aligned}$ The $4^{\text{th}}$ term is $29$.